Let m and n be positive integers. First consider the case where m ≠ n. By the product identity 
sin(x) sin(y) = 
1
2
 

cos(x − y) − cos(x + y)

,
 the integral can be rewritten as follows.
𝜋

−𝜋
 sin(mx) sin(nx) dx	 = 	
𝜋

−𝜋
 
1
2
 

cos 

 
  
Correct: Your answer is correct.
 

 − cos(mx + nx)

 
 = 	
𝜋

−𝜋
 
1
2
 

cos 


 
  
Correct: Your answer is correct.
 

 x

 − cos((m + n)x)

 
 = 	
1
2
 

 
  


We start with the given product identity:

\[
\sin(mx) \sin(nx) = \frac{1}{2} \left( \cos(x - y) - \cos(x + y) \right)
\]

### Step 1: Apply the identity to the integral

Using the identity for \(\sin(mx)\sin(nx)\), we rewrite the integral:

\[
\int_{-\pi}^{\pi} \sin(mx) \sin(nx) \, dx = \int_{-\pi}^{\pi} \frac{1}{2} \left( \cos((m - n)x) - \cos((m + n)x) \right) \, dx
\]

### Step 2: Split the integral

Now, we separate the integrals for \(\cos((m - n)x)\) and \(\cos((m + n)x)\):

\[
\frac{1}{2} \int_{-\pi}^{\pi} \cos((m - n)x) \, dx - \frac{1}{2} \int_{-\pi}^{\pi} \cos((m + n)x) \, dx
\]

### Step 3: Evaluate the integrals

Recall that the integral of \(\cos(kx)\) over the interval \([- \pi, \pi]\) for any non-zero integer \(k\) is zero:

\[
\int_{-\pi}^{\pi} \cos(kx) \, dx = 0 \quad \text{if } k \neq 0
\]

So, for \(m \neq n\):

\[
\int_{-\pi}^{\pi} \cos((m - n)x) \, dx = 0 \quad \text{and} \quad \int_{-\pi}^{\pi} \cos((m + n)x) \, dx = 0
\]

### Step 4: Final answer for \(m \neq n\)

Thus, the entire integral evaluates to zero:

\[
\int_{-\pi}^{\pi} \sin(mx) \sin(nx) \, dx = 0 \quad \text{if } m \neq n
\]

### Step 5: Case where \(m = n\)

If \(m = n\), we return to the original integral:

\[
\int_{-\pi}^{\pi} \sin(mx) \sin(nx) \, dx = \int_{-\pi}^{\pi} \sin^2(mx) \, dx
\]

Using the identity \(\sin^2(x) = \frac{1}{2}(1 - \cos(2x))\), we get:

\[
\int_{-\pi}^{\pi} \sin^2(mx) \, dx = \int_{-\pi}^{\pi} \frac{1}{2} (1 - \cos(2mx)) \, dx
\]

The integral of \(\cos(2mx)\) over \([- \pi, \pi]\) is zero, so we are left with:

\[
\int_{-\pi}^{\pi} \frac{1}{2} (1 - \cos(2mx)) \, dx = \frac{1}{2} \int_{-\pi}^{\pi} 1 \, dx = \frac{1}{2} \times 2\pi = \pi
\]

### Final result for \(m = n\):

\[
\int_{-\pi}^{\pi} \sin(mx) \sin(nx) \, dx = \pi \quad \text{if } m = n
\]

---

In summary, we have:

\[
\int_{-\pi}^{\pi} \sin(mx) \sin(nx) \, dx =
\begin{cases}
0, & \text{if } m \neq n \\
\pi, & \text{if } m = n
\end{cases}
\]

Would you like more details, or do you have any other questions?

---

### 5 Follow-Up Questions:
1. How can this result be applied in Fourier series analysis?
2. What is the geometric interpretation of the orthogonality of sine functions?
3. How would the result change if we integrated over a different interval, such as \([0, 2\pi]\)?
4. Can you generalize this approach to functions like \(\sin(ax)\sin(bx)\)?
5. How would the integral look for cosine functions instead of sine?

### Tip: Remember that the orthogonality of sine and cosine functions on symmetric intervals like \([- \pi, \pi]\) is a key property used in Fourier series decomposition.

Let m and n be positive integers. First consider the case where m ≠ n. By the product identity sin(x) sin(y) = (1/2) [cos(x − y) − cos(x + y)], the integral can be rewritten as follows. ∫ from -π to π sin(mx) sin(nx) dx = (1/2) [∫ from -π to π cos(mx − nx) dx − ∫ from -π to π cos(mx + nx) dx].

Explore the integral of sin(mx)sin(nx) over the interval [-π, π], using trigonometric identities and the orthogonality of sine functions. This detailed solution explains key concepts in trigonometry and integral calculus, suitable for undergraduate calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation